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## New answers tagged continued-fractions

1

The answer to Q1 is yes. About Q2, I don't know yet. As usual, I suppose that $q=\exp(2\pi\mathrm{i}\tau)$ for $\tau\in\mathbb{H}$ (complex upper half plane) and define $$q_n = \exp\frac{2\pi\mathrm{i}\tau}{n}$$ Thus we can consider Theta functions to be functions of $\tau$. I write $\vartheta_k(0\mid\tau)$ instead of $\vartheta_k(0,q_2)$. This is mostly to ...

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Chrystal's Algebra Volume Two. See this note for a link to an online electronic copy: http://recursed.blogspot.com/2009/12/chrystals-algebra-is-available.html For something more recent, try Doug Hensley's Continued Fractions.

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I refer to your claim with the sign of $q$ adjusted so that it reads $$\small(1+q^{2}+q^{6}+q^{12}+q^{20}+q^{30}+\cdots)^{2} =\cfrac{1}{1-q+{\cfrac{q\,(1-q)^2}{1-q^3+\cfrac{q^2(1-q^2)^2} {1-q^5+\cfrac{q^3(1-q^3)^2}{1-q^7+\cdots}}}}}$$ Given a related answer introducing a continued fraction formula by Ramanujan with parameters $a,b,q$ and making use of some ...

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I write $q_n = \exp\frac{2\pi\mathrm{i}\tau}{n}$, thus $q_n^n=q$. In a related answer and another one, I used a formula by Ramanujan, proved by Adiga et al. (1985): $$\small\frac{(-a;q)_\infty\,(-b;q)_\infty - (a;q)_\infty\,(b;q)_\infty} {(-a;q)_\infty\,(-b;q)_\infty + (a;q)_\infty\,(b;q)_\infty} = ... 3 In a related answer, I used a formula by Ramanujan, proved by Adiga et al. (1985):$$\small\frac{(-a;q)_\infty\,(-b;q)_\infty - (a;q)_\infty\,(b;q)_\infty} {(-a;q)_\infty\,(-b;q)_\infty + (a;q)_\infty\,(b;q)_\infty} = \cfrac{a+b}{1-q+\cfrac{(a+bq)(aq+b)}{1-q^3+\cfrac{q\,(a+bq^2)(aq^2+b)} {1-q^5+\cfrac{q^2(a+bq^3)(aq^3+b)}{1-q^7+\cdots}}}}\tag{*}$$Applying ... 2 For those interested, here is an outline of a proof. To prepare the ground, let \mathbb{H} be the complex upper half plane, \tau\in\mathbb{H} and$$\begin{align} q_n &\stackrel{\text{def}}{=} \exp\frac{2\pi\mathrm{i}\tau}{n} \\ q &\stackrel{\text{def}}{=} q_1 \end{align}$$so I can write q_2 instead of q^{1/2} etc. Furthermore, let us ... 0 take 5+6/[5+6/5...]=x then we can write , [5+6/x]=x 5x+6=x^2 x^2-5x-6=0 (1) x=-1 and x=6 are the roots of (1) since the sequence of partial sum of above series is of positvie terms limit cannot be -1[because sequence of +ve terms must converge to a limit >0] hence limit must be equals to '6' hence above sum is equals to '6' 0 I really dislike the equality$$\sum _{n=1}^{\infty }n=-\frac {1} {12}.$$It is much better to write it as$$\zeta(-1)=-\frac1{12},$$where \zeta is Riemann's zeta function. And$$\sum _{n=1}^{\infty }n=+\infty$$is absolutely true. And for the original question, the method is valid only when the expression has a value (i.e. a certain sequence has a ... 1 The key is in the dots. It is tough to get a handle on an infinite thing, so usually, it would be a process: Start with x_1=5 Calculate x_2=5+6/x_1 Calculate x_3=5+6/x_2 Carry on; what is the limit of x_n? Suppose x_n=6+y_n. Then$$y_{n+1}=x_{n+1}-6\\ =6/x_n-1\\ =6/(6+y_n)-1\\ =-y_n/(6+y_n)$$Since -1\leq y_1\leq1, we ... 2 Note that you can take any sequences a_n\to\alpha and b_n\to\beta in order to obtain a sequence converging to \alpha+\beta (or \alpha\cdot\beta); you don't need continued fractions in order to obtain a representant sequence for \alpha+\beta. Now in your approach any real number \alpha, rational or irrational, has a certain unique "normal form": ... 3 See: Rieger, “A new approach to the real numbers (motivated by continued fractions),” Abhandlungen der Braunschweigischen Wissenschaftlichen Gesellschaft, vol. 33, pp. 205–217, 1982. The author gives a construction of the reals using continued fractions. A survey of this construction appears in this article. I'm not sure how this helps though in studying ... 3 From$$y=\frac{1}{1+y} \tag{1}$$the resulting quadratic equation should be$$y^2+y−1=0 \tag{2}$$with the two solutions being y=\frac{−1\pm\sqrt5}{2}. It turns out that these values of y are the fixed points of the function f:\mathbb{R}\to\mathbb{R} where$$f(x)=\frac{1}{1+x} \tag{3}$$The continued fraction converges to whatever value the ... 4 (A half-answer.) Let |q|<1 and,$$p = q^m,\quad \alpha = \pm q^{-n},\quad \beta = \pm q^n,\quad \alpha\beta = 1$$We propose that,$$\small C_{m,n}(q)=\prod_{k=1}^\infty\frac{(1-\alpha\, p^{4k})(1-\beta\, p^{4k-4})}{(1-\alpha\, p^{4k-2})(1-\beta\, p^{4k-2})} \overset{\color{red}{?}}=\cfrac{(1-\beta\,p^0)}{ 1-p+\cfrac{p(1-\alpha\, p)(1-\beta\, ...

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(A partial answer.) Compare the three cfracs of similar form, $$H(q)=\frac{q^{1/2}\,\vartheta_3(0,q^2)}{\vartheta_2(0,q^2)}=\small\cfrac{(1+q^2)}{1-q+\cfrac{q(1+q^{-1})(1+q^3)}{1-q^3+\cfrac{q^2(1+q^0)(1+q^4)}{1-q^5+\cfrac{q^3(1+q)(1+q^5)}{1-q^7+\cfrac{q^4(1+q^2)(1+q^6)}{1-q^9+\ddots}}}}}\tag1$$ $$U(-q)\; =\; ... 3 As you suggest, writing x in continued fractions x=[x_0,x_1,\ldots x_k] determines univocally the color of x: if a_0+a_1+\ldots +a_k is odd then x is yellows, otherwise is green. It remains to see if this coloring satisfies the two rules: i) if x=[x_0,x_1,\ldots x_k], then 1/x=[0,x_0,x_1,\ldots x_k], so they have the same color ii)if ... 0 I think you can construct something form$$\sqrt{a}=1+\sqrt{a}-1=1+\frac{(\sqrt{a}-1)(\sqrt{a}+1)}{1\cdot(\sqrt{a}+1)}=1+\frac{a-1}{1+\sqrt{a}}$$This equation is self referential. So if you plug in the the expression on the right hand side of the equation for \sqrt{a} in the denominator you get. ... 1 Too long for a comment first let's assume the following identity to be true$$\frac{1}{q}\sum_{n=1}^\infty \frac{q^n}{1-q^{2n-1}} = \cfrac{1}{1-q-\cfrac{q(1-q)^2}{1-q^3-\cfrac{q^2(1-q^2)^2}{1-q^5-\cfrac{q^3(1-q^3)^2}{1-q^7-\ddots}}}}$$Now multiplying both sides by (1-q) and letting q\rightarrow1, yields$$\lim_{q\rightarrow 1} ...

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